Algebraic Curves Lecture 22

\(d \lt 2r\) means you’ve got a non-special line bundle. Not sure where this showed up last time as I wasn’t there.

Suppose \(C\subset \mathbb P^r\) is embedded by \(\mathcal O_{\mathbb P^1}(r)\),

\begin{align*} [s:t] \mapsto [s^r:s^{r-1}t:...:t^r]. \end{align*}

Remember:

• such maps are characterized as non-degenerate curves of degree \(r\) in \(\mathbb P^r\)
• any \(r+1\) points on \(C\) span \(\mathbb P^r\).
• \(\varphi_{K_X}(X)\) is a rational normal curve if \(X\) is hyperelliptic.

Set

\begin{align*} M = \begin{pmatrix} x_0 & x_1 & x_2 & \dots & x_{r-1} \\ x_1 & x_2 & x_3 & \dots & x_r \end{pmatrix}. \end{align*}

The ideal of \(C\) is defined by \(2\times 2\) minors of \(M\). There are \(\binom{r}{2}\) linearly independent quadrics cutting out \(C\).

\begin{align*} \varphi:H^0(\mathbb P^r, \mathcal O_{\mathbb P^r}(2)) \to H^0(C, \mathcal O_C(2)). \end{align*}

and

\begin{align*} \dim(\operatorname{img}\varphi) &= \binom{r+2}{2} - \binom{r}{2} \\ &= \frac{(r+2)(r+1) - r(r-1)}{2} \\ &= \frac{r^2 + 3r + 2 - r^2 + r}{2} \\ &= 2r + 1 = h^0(C,\mathcal O_C(2)) \end{align*}

which implies that \(\varphi\) is surjective which happens if and only if \(C\) is 2-normal.

Proof

Assume \(C\) is \(0\), \(1\), \(2\)-normal. We need to show that \(C\) is \(\ell\) normal for any \(\ell > 3\). Consider the exact sequence:

\begin{align*} 0 \to \mathcal O_C(1) \xrightarrow{\cdot H} \mathcal O_C(2) \to \mathcal O_{C\cap H}(2)\to 0. \end{align*}

As is standard, take the long exact sequence on cohomology

\begin{align*} H^{0}(\mathbb P^r, \mathcal O_{\mathbb P^r}(2)) \to H^0(C, \mathcal O_C(2)) \to H^0(C\cap H, \mathcal O_{C\cap H}(2)) \to H^1(C, \mathcal O_C(1)). \end{align*}

This means the points in \(C\cap H\) impose independent conditions on quadrics in \(\mathbb P^r\) which in turn implies the points also impose independent conditions on all hypersurfaces of degree \(\ell \geq 2\).

Now consider the exact sequence of the ideal sheaf \(\mathcal I_{C\cap H}\) given by

\begin{align*} 0 \to \mathcal I_{C\cap H}(\ell) \to \mathcal O_{\mathbb P^r}(\ell) \to \mathcal O_{C\cap H}(\ell) \to 0. \end{align*}

Then

\begin{align*} H^1(C, \mathcal I_{C\cap H}(\ell)) = 0 \end{align*}

for all \(\ell\geq 2\), since \(\mathcal O_{\mathbb P^r}(\ell) \to \mathcal O_{C\cap H}(\ell)\) is surjective on global sections whenever \(\ell \geq 2\). We’d like to show that \(H^1(C, \mathcal I_{C\cap H}(1) = 1\) too. We have that

\begin{align*} 0 \to \mathcal I_{C\cap H}(\ell - 1) \to \mathcal I_{C\cap H}(\ell) \to \mathcal I_{C\cap H}^H(\ell) \to 0 \end{align*}

where \(\mathcal I_{C\cap H}\) is the ideal sheaf of \(C\cap H\) inside of \(\mathcal P^r\) and \(\mathcal I_{C\cap H}^H(\ell)\) is the ideal sheaf of \(C\cap H\) inside of \(H\cong \mathbb P^{r-1}\). This exact sequence comes from the fact that…I’m not sure. Will need to consult the notes.

Thus, rational normal curves are projectively normal.

The rational normal scrolls will be surfaces which contain our extremal curves. Fix \(a_1\) and \(a_2\) such that \(a_1\geq a_2\), \(a_1\geq 0\) \(a_2\geq 0\) and \(a_1 + a_2 > r - 1\). Fix subspaces \(L_1\) and \(L_2\) such that \(L_1\) and \(L_2\) span \(\mathbb P^r\) and \(\dim(L_i) = a_i\). Let \(C_i \subseteq C_i \subseteq L_i\) be a rational normal curve and fix an isomorphism \(\varphi:C_1\to C_2\). Then

\begin{align*} X_{a_1, a_2} = \bigcup_{p\in C_1} \{\text{line connecting } P \text{ to } \varphi(P)\}. \end{align*}

Example:

• Take \(r = 2\), then \(X_{0,1}\cong \mathbb P^2\).
• Take \(r = 3\). Then \(X_{0,2} - \) cone over a conic, \(X_{1,1} \cong \mathbb P^1\times \mathbb P^1\).
• Take \(r = 4\). Then \(X_{1,2} \operatorname{BL}_p\mathbb P^2\).

An Observation: Up to linear automorphisms of \(\mathbb P^r\) this construction does not depend on the choices made other than \(a_1\) and \(a_2\).

Proof

1. First, \(\operatorname{span}(c_i) = L_i\) and \(\operatorname{span}(L_1\cup L_2) = \mathbb P^r\). Any linear subspace \(H\supset X_{a_1, a_2}\) needs to contain the \(C_i \implies L_i \subset H\), hence \(\operatorname{span}(L_1\cup L_2) \subset H \implies H = \mathbb P^r\).
2. \(X_{a_1,a_2} \cap H = C_2 + \sum_{i=1}^{a_1}L_i\). Thus the degree of \(X_{a_1,a_2}\) is equal to the degree of \(X_{a_1,a_2}\cap H\) which is \(a_1 + a_2\).
3. \(X_{a_1, a_2} \cap H_p\). For any \(p\), I can choose \(H_p\) such that \(X_{a_1,a_2} \cap H_p\) is smooth at \(p\) as long as \(a_1\neq 0\). Then

\begin{align*} \operatorname{deg}(H\cap X_{a_1,a_2}) = r-1 \end{align*}

which implies \(X_{a_1,a_2} \cap H\subset H\) is a rational normal curve if smooth.

What about the Picard group of \(X_{a_1,a_2}\)? This is generated freely by the class \(L\) of a line in the ruling and by \(H\) the class of a hyperplane section. Note that you can write \(H = C_1 + a_2 \cdot L = C_2 + a_1 \cdot L\), so you can choose the class of one of the curves instead of \(H\) if you prefer. Some intersection pairings:

• \(L.L = 0\)
• \(L.H = 1\)
• \(H.H = a_1 + a_2 = r - 1\)

Remarks: Remember

\begin{align*} M = \begin{pmatrix} x_0 & x_1 & x_2 & \dots & x_{r-1} \\ x_1 & x_2 & x_3 & \dots & x_r \end{pmatrix}. \end{align*}

Set

\begin{align*} M' = \begin{pmatrix} x_0 & x_1 & x_2 & \dots & x_{a_1-1} & y_0 & y_1 & \dots & y_{a_2 - 1} \\ x_1 & x_2 & x_3 & \dots & x_{a_1} & y_0 & y_1 & \dots & y_{a_2} \end{pmatrix}. \end{align*}

1. \(X_{a_1, a_2}\) will be cut out by \(2\times 2\) minors of \(M'\), \(\binom{r-1}{2}\) quadrics.
2. The surface \(X_{a_1,a_2}\) is projectively normal: \[0 \to I_{X_{a_1,a_2}} \xrightarrow{\cdot H} I_{X_{a_{1},a_2}}(\ell) \to I_{X_{a_1,a_2}\cap H}^H(\ell) \to 0.\]
3. \(X_{a_1,a_2}\cong \mathbb P(\mathcal O_{\mathbb P^1}(a_1)\oplus \mathcal O_{\mathbb P^1}(a_2))\cong \mathbb F_{|a_2 - a_1|}\) where \(\mathbb F_{|a_2 - a_1|}\) is a Hirzebruch surface.